what it is

Power series - Wikipedia

power series

#card/reverse it's actually a type of (simpler) taylor series. weird but true \sum_{n=0}^{\infty} a_{n}(x-c)^{n}=a_{0}+a_{1}(x-c)+a_{2}(x-c)^{2}+\ldots

mclaurin

\sum_{n=0}^{\infty} a_{n} x^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots

3 possibilities for the value of x

x=a Ratio test gives \infty
(-\infty,\infty), Ratio test = 0
find R>0 (radius of convergence) where |x-a|<R converges. Ratio test gives R where -R+a<x<R+a

intervals and derivatives

$f'(x)= c_{1}+2c_{2}(x-a)+3c_{3}(x-a)^2...=\sum\limits_{n=1}^{\infty}nc_n(x-a)^{n-1}=\sum\limits_{n=0}^{\infty}(n+1)c_{n}(x-a)^{n}$ \int f(x)dx=C + c_{o}(x-a)+c_1\frac{(x-a)^2}{2}+c_2\frac{(x-a)^3}{3}...=C+\sum\limits_{n=0}^{\infty}c_n\frac{(x-a)^{n+1}}{n+1}

use to solve integrals

any part of the integral that might be approximated by a power series can just be swapped out by its power series
integrate and sum operations can be swapped around
power series of a function is a geometric series, so its sum can be swapped in. \sum\limits_{n=0}^{\infty}a_{n}(x-a_{n})^{n}=a\frac{1-r^{n}}{1-r}

1.2 KiB Raw Blame History